[tex]\it \Delta ABC-dr, \hat A=90^o,\ \stackrel{TP}{\Longrightarrow}\ AC^2=BC^2-AB^2 \Rightarrow AC^2=20^2-16^2 \Rightarrow \\ \\ \Rightarrow AC^2=(20-16)(20+16)=4\cdot36=2^2\cdot6^2 \Rightarrow AC=2\cdot6=12\ cm[/tex]
[tex]\it AD=\dfrac{AB\cdot AC}{BC}=\dfrac{16\cdot12^{(2}}{20}=\dfrac{16\cdot6}{10}=\dfrac{96}{10}=9,6\ cm\\ \\ \\ \Delta DAB-dr,\ \hat D=90^o,\ \stackrel{TP}{\Longrightarrow}\ BD^2=AB^2-AD^2 \Rightarrow BD^2=16^2-9,6^2 \Rightarrow \\ \\ \\ \Rightarrow BD^2=(16-9,6)(16+9,6)=6,4\cdot25,6=\dfrac{64\cdot256}{100}=\dfrac{8^2\cdot16^2}{10^2} \Rightarrow \\ \\ \\ \Rightarrow BD=\dfrac{8\cdot16}{10}=\dfrac{128}{10}=12,8\ cm[/tex]
[tex]\it DC = BC-BD=20-12,8=7,2\ cm\\ \\ \mathcal{A}=\dfrac{c_1\cdot c_2}{2}=\dfrac{16\cdot12}{2}=16\cdot6=96\ cm^2[/tex]
